Question: A game at the carnival offers these odds: you get to roll a ten-sided die, and if you roll more than a 6, you make $6$ dollars. Unfortunately, if you roll anything else, you lose $9$ dollars. How much money do you expect to make (or lose) playing this game?
Solution: The expected value of an event (like playing this game) is the average of the values of each outcome. Since some outcomes are more likely than others (sometimes), we weight the value of each outcome according to its probability to get an accurate idea of what value to expect. There are two events that can happen in this game: either you roll more than a 6, or you don't. So, the expected value will look like this: $E = $ (money gained when you roll more than a 6) $\cdot$ (probability of rolling more than a 6) $+$ (money gained when you don't roll more than a 6) $\cdot$ (probability of not rolling more than a 6). The money you gain when you win is $$6$ . The probability of winning is the probability that you roll more than a 6. This probability is the number of winning outcomes divided by the total number of outcomes, $\dfrac{4}{10}$ The money you gain when you lose is $$ -9$ (since you actually lose money). The probability that you lose is the probability that you don't roll more than a 6. This probability must be $1 - \dfrac{4}{10} = \dfrac{6}{10}$ So, if we take the average of the amount of money you make on each outcome, weighted by how probable each outcome is, we get the expected amount of money you will make: $(6\cdot\dfrac{4}{10}) + (-9\cdot\dfrac{6}{10}) = -\dfrac{30}{10} = -$3.00. $